HDOJ 4417 Super Mario 線段樹 單點更新 成段查詢

//HDOJ 4417 Super Mario 線段樹 轉化爲單點更新 成段查詢

/*
題意：查詢區間內小於等於某數的數的總和

思路：離線查詢，先記錄所有的詢問並排序，然後一邊刪一邊查，查之前把不滿足的刪掉
	  就轉化爲簡單的單點更新，成段求和。
*/

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define N 100005

int T,n,m,cnt,ca = 1;
int sum[N<<2],ans[N];

struct Node{
	int x;
	int id;
}s[N];

struct node{
	int from;
	int to;
	int k;
	int id;
}q[N];

int cmp(const void *a,const void *b){
	return (*(Node *)b).x - (*(Node *)a).x;
}

int cmp1(const void *a,const void *b){
	return (*(node *)b).k - (*(node *)a).k;
}

void Pushup(int rt){
	sum[rt] = sum[rt<<1]+sum[rt<<1|1];
}

void Debug(int n){
	int i;
	for(i = 1; i <= n; ++i)
		printf("sum[%d] : %d\n",i,sum[i]);
}

void Build(int rt,int l,int r){
	if(l == r){
		scanf("%d",&s[cnt].x);
		s[cnt].id = cnt++;
		sum[rt] = 1;
		return ;
	}
	int mid = (l + r) >> 1;
	Build(lson);
	Build(rson);
	Pushup(rt);
}

void Update(int rt,int l,int r,int x,int c){
	if(l == r){
		sum[rt] += c;
		return ;
	}
	int mid = (r + l) >> 1;
	if(x <= mid)	Update(lson,x,c);
	else			Update(rson,x,c);
	Pushup(rt);
}

int Query(int rt,int l,int r,int L,int R){
	if(L <= l && R >= r){
		return sum[rt];
	}
	int res = 0;
	int mid = (l + r) >> 1;
	if(L <= mid) res += Query(lson,L,R);
	if(R > mid ) res += Query(rson,L,R);
	return res;
}

void Init(){
	int i;
	cnt = 1;
	scanf("%d %d",&n,&m);
	Build(1,1,n);
	qsort(s+1,n,sizeof(s[0]),cmp);
	for(i = 1; i <= m; ++i){
		scanf("%d %d %d",&q[i].from,&q[i].to,&q[i].k);
		q[i].id = i;
	}
	qsort(q+1,m,sizeof(q[0]),cmp1);
}

void Solve(){
	int i,j = 1;
	memset(ans,0,sizeof(ans));
	for(i = 1; i <= n; ++i){
		while(s[i].x <= q[j].k){
			ans[q[j].id] = Query(1,1,n,q[j].from+1,q[j].to+1);
			++j;
			if(j > m)	break;
		}
		if(s[i].x > q[j].k){
			Update(1,1,n,s[i].id,-1);	
		}
	}
}

void Print(){
	int i;
	printf("Case %d:\n",ca++);
	for(i = 1; i <= m; ++i)
		printf("%d\n",ans[i]);
}

int main(){
	scanf("%d",&T);
	while(T--){
		Init();
		Solve();
		Print();
	}
	return 0;
}